When higher voltage pulses are used to drive the motor, the current rise can be increased. Voltage acts as a pressure to force current into the windings at a faster rate. When the current rises at a steeper rate, it can achieve the rated current value in a shorter amount of time. So raising the voltage provides more current than lower voltage especially as the pulse frequency increases. Higher current at a high pulse rate translates to higher torque at high speed. There are a couple types of drives used to achieve higher speeds with stepper motors.
L/R drives
An L/R driver increases the driver voltage while keeping the current flowing to the motor at or below the rated current. If one increases the current above the rated current, the motor will get very hot as the heating power is I2R. Suppose there is a motor rated for 6 VDC, 1 A, and 6 Ω . If the motor is supplied with only the rated voltage of 6V it will have only minimal performance (6VDC/ 6Ω = 1 AMP). However, if an external 6 Ω external resistor is placed in series with the windings then the total resistance per phase increases to 12 Ω . With the 6VDC signal, the motor will only receive 0.5 amps. Therefore, with the 6 Ω external resistor, we can now increase the driver voltage to 12 VDC, since I=12V/12 Ω = 1 amp which is the rated current. Since the circuit inductance, L, is essentially the same and the circuit resistance, L, has doubled the circuit time constant L/R is decreased and current will rise faster in the windings. So the torque vs. speed curve will be pushed out at higher speeds. Further, suppose we increase the external resistance to 18 Ω . The total resistance is now 24 Ω and with a 12 VDC power supply, the motor will only get 0.5 amps. Therefore, we can increase the driver voltage to 24 VDC, since I=24 D/24 Ω = 1 amp which, again, is the rated current. Because the circuit time constant, L/R, is further decreased, the torque/speed curve will be pushed out even more. The L/R driver is effective in that it increases the high-speed torque. It is also relatively simple to implement. However, it is inefficient as the putting current through the added resistance increases the power loss.